Date: Jun 16, 2013 4:38 AM
Author: Bruno Luong
Subject: Re: eigenvector
"Remus " <remusac@yahoo.com> wrote in message <kpighh$qj3$1@newscl01ah.mathworks.com>...

>

> Hi guys,

> I'm also looking at this problem. To answer you question John, yes I know that 1p is an Eigenvector of A. Let A be a matrix which has sum of all rows = 0, then 1p (where p=dim(a)) is an eigenvector of A.

> For Example let A = [0 0 0; -1 1 0; -1 0 1]. the comand [V,D]=eig(A) returns the following eigenvectors:

> V = [0 0 .5774; 1 0 .5774; 0 1 .5774], which is correct but as posted in the thread it is not normalized to 1.

It *is* normalized to 1

>> A = [0 0 0; -1 1 0; -1 0 1]

A =

0 0 0

-1 1 0

-1 0 1

>> [V,D]=eig(A)

V =

0 0 0.5774

0 1.0000 0.5774

1.0000 0 0.5774

D =

1 0 0

0 1 0

0 0 0

>> sqrt(sum(V.^2,1)) % compute l2-norm of 3 eigen vectors

ans =

1 1 1

>>

% Bruno